Test Index

UPSC NDA I 2026 Mathematics Paper

© examsnet.com

Question : 29 of 120

Marks: +1, -0

If

\cos \theta=\frac{1}{3}

, then what is the value of

\sin \left(\frac{\boldsymbol{\theta}}{\mathbf{2}}\right) \boldsymbol{\operatorname { s i n }}\left(\frac{\mathbf{3} \boldsymbol{\theta}}{\mathbf{2}}\right) ?

$\frac{5}{9}$
$\frac{7}{9}$
$\frac{10}{9}$
$\frac{11}{9}$

Solution:

Concept:

Use the product-to-sum formula:

2\sin A \sin B = \cos(A-B) - \cos(A+B)

.

Explanation:

Let

A = \frac{\theta}{2}

and

B = \frac{3\theta}{2}

.

Then,

\sin\frac{\theta}{2}\sin\frac{3\theta}{2} = \frac{1}{2}\left[\cos\left(\frac{\theta}{2}-\frac{3\theta}{2}\right) - \cos\left(\frac{\theta}{2}+\frac{3\theta}{2}\right)\right]

= \frac{1}{2}\left[\cos(-\theta) - \cos 2\theta\right] = \frac{1}{2}\left[\cos\theta - \cos 2\theta\right]

Given

\cos\theta = \frac{1}{3}

,

\cos 2\theta = 2\cos^2\theta - 1 = 2\left(\frac{1}{3}\right)^2 - 1 = \frac{2}{9}-1 = -\frac{7}{9}

So,

\frac{1}{2}\left[\frac{1}{3} - \left(-\frac{7}{9}\right)\right] = \frac{1}{2}\left[\frac{3}{9}+\frac{7}{9}\right] = \frac{1}{2}\cdot\frac{10}{9} = \frac{5}{9}

Answer:

\frac{5}{9}

, i.e. Option A.

© examsnet.com

Go to Question:

Prev Question

Next Question